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For positive integers $m,n$, let $f(m,n)$ denote the number of positive integers which are both a multiple of $m$ and a factor of $n$. Find $\displaystyle \sum_{i=1}^\infty\left(\frac 1 {i^2}\sum_{j=1}^if(j,i)\right)$. Hint: $\displaystyle\sum_{i=1}^\infty\frac 1 {i^2}=\frac{\pi^2} 6$.

This is a question from a maths contest. I have no idea to solve it. Do anyone have any idea? Thank you.

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What contest is this by the way? – Ishan Banerjee Mar 26 '13 at 9:47
It's just a small regional contest. – ᴊ ᴀ s ᴏ ɴ Mar 26 '13 at 9:49
up vote 4 down vote accepted

Note that \begin{align} f(m , n)= \begin{cases} 0& \text{if} \, m \nmid n \\ d\left(\frac{n}{m}\right) & \text{if} \, m \mid n \end{cases} \end{align}

where $d(n)=\sum_{d \mid n}{1}$ denotes the number of divisors of $n$. Thus

\begin{align} \sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j=1}^{i}{f(j,i)}\right)}& =\sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j \mid i}{d\left(\frac{i}{j}\right)}\right)} \\ & =\sum_{i=1}^{\infty}{\left(\frac 1 {i^2}\sum_{j \mid i}{d(j)}\right)} \\ & =\sum_{j=1}^{\infty}{d(j)\sum_{j \mid i, i \geq 1}{\frac{1}{i^2}}} \\ & =\sum_{j=1}^{\infty}{d(j)\sum_{k=1}^{\infty}{\frac{1}{(jk)^2}}} \\ & =\sum_{j=1}^{\infty}{\frac{d(j)}{j^2}\sum_{k=1}^{\infty}{\frac{1}{k^2}}} \\ & =\frac{\pi^2}{6}\sum_{j=1}^{\infty}{\frac{d(j)}{j^2}} \\ & =\frac{\pi^2}{6}\sum_{j=1}^{\infty}{\frac{1}{j^2}\sum_{d|j}{1}} \\ & =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\sum_{d \mid j, j \geq 1}{\frac{1}{j^2}}} \\ & =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\sum_{l=1}^{\infty}{\frac{1}{(dl)^2}}} \\ & =\frac{\pi^2}{6}\sum_{d=1}^{\infty}{\frac{1}{d^2}\sum_{l=1}^{\infty}{\frac{1}{l^2}}} \\ & =\frac{\pi^4}{36}\sum_{d=1}^{\infty}{\frac{1}{d^2}} \\ & =\frac{\pi^6}{216} \end{align}

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